<p>For a certain tire, the manufacturere recommends that the air pressure be between 30 and 34 pounds per square inch. If x pounds per square inch is the air pressure in a tire that meets this recommendation, which of the following represents all possible values of x?</p>
<p>a) |x-32| < 2
b) |x+32| < 2
c) |x-32| > 2
d) |x+32| > 2
e) |x-32| = 2</p>
<p>A.</p>
<p>10char</p>
<p>Why can’t it be B as well?</p>
<p>Hey, A is right. Can you help me understand why? I just plugged in some values both inside and outside the bounds and narrowed it down to A. I’d like to get some intuition, though.</p>
<p>It’s A because for B you have |x+32| < 2 and you know that your air pressure is going to be between 30 and 34 so no matter the value of x, |x+32| is ALWAYS going to be > 2. Do you understand???</p>
<p>um… i said intuition. I already plugged in and found.</p>
<p>Oh duh thanks FrenchieGirl, I misread the question.</p>
<p>In strict math:</p>
<p>|x-32| < 2 ==> -2 < (x-32) < 2 ==> 30 < x < 34
|x+32| > 2 ==> (x+32) < -2 or (x+32) > 2 ==> x < -34 or x > -30</p>
<p>you’re welcome!! ;-)</p>
<p>Since this kind of question does come up…</p>
<p>When you are given a range of values that x must fall in, say 70 to 90, another way to think of it is to find the size of the range (in this case, 90 -70 = 20), cut that in half (so now you have 10) and then say that x has to be within that distance of the midpoint of the range. </p>
<p>So 70 < x < 90 is equivalent to |(x-80)| < 10 …saying that when you subtract, you’ll get a difference of less than 10, greater than -10.</p>
<p>In OP’s problem, the range is 30 to 34…which is a range of 4, so we want x to be within 2 of the midpoint so |x-32|<2.</p>